eureka math 3rd grade lesson 9 homework

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3rd grade (Eureka Math/EngageNY)

Unit 1: module 1: properties of multiplication and division and solving problems with units of 2–5 and 10, unit 2: module 2: place value and problem solving with units of measure, unit 3: module 3: multiplication and division with units of 0, 1, 6–9, and multiples of 10, unit 4: module 4: multiplication and area, unit 5: module 5: fractions as numbers on the number line, unit 6: module 6: collecting and displaying data, unit 7: module 7: geometry and measurement word problems.

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Eureka Lesson 9 Homework Grade 3

Eureka Lesson 9 Homework Grade 3 - Displaying top 8 worksheets found for this concept.

Some of the worksheets for this concept are Eureka math homework helper 20152016 grade 2 module 4, Eureka math homework helper 20152016 grade 2 module 3, Eureka math homework helper 20152016 grade 1 module 1, Eureka math homework helper 20152016 grade 5, Eureka math homework helper 20152016 grade 6 module 2, Eureka math homework helper 20152016 grade 3 module 1, Table of contents grade 2 module 3, Louisiana guide to implementing eureka math grade 3.

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1. Eureka Math Homework Helper 20152016 Grade 2 Module 4

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Interactive Digital Curriculum   - The Eureka Navigator is a complete PreK-12 curriculum, that includes embedded PD videos demonstrating key instructional practices and an interactive map of the modules for easy navigation.

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Topic outline, grade 3 module 7, geometry and measurement word problems.

Topic A: Solving Word Problems

Topic B: Attributes of Two-Dimensional Figures

Topic c: problem solving with perimeter, mid-module review and assessments, topic d: recording perimeter and area data on line plots, topic e: problem solving with perimeter and area, topic f: year in review, end-of-module review and assessments.

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Eureka Math Grade 4 Module 3 Lesson 9 Answer Key

Engage ny eureka math 4th grade module 3 lesson 9 answer key, eureka math grade 4 module 3 lesson 9 problem set answer key.

Explanation: In standard algorithm we add same time of multiplying as 1 251 X 3 753 , here 3 X 1 ones = 3 ones then 3 x 5 tens = 15 tens we write 5 at tens place and take 1 to hundred place and 3 X 2 hundreds = 6 hundreds, now adding 1 hundred to 6 hundreds we get 7 hundreds , so 3 X 251 = 753.

Explanation: In standard algorithm we add same time of multiplying as 1, 3 135 X 6 810 , here 6 X 5 ones = 30 ones, we write 0 at ones place and take 3 to tens places then 6 x 3 tens = 18 tens + 3 tens = 21 tens, we write 1 at tens place and take 2 to hundred place and 6 X 1 hundred = 6 hundreds, now adding 2 hundreds to 6 hundreds we get 8 hundreds , so 6 X 135 = 810.

Explanation: In standard algorithm we add same time of multiplying as  3 304 X 9 2,736 ,here 9 X 4 ones = 36 ones, we write 6 at ones place and take 3 to tens places then 9 x 0 ten = 0 ten + 3 tens = 3 tens, 9 X 3 hundreds = 27 hundreds, now we write 7 at hundreds place and 2 at thousands place as shown above, So 9 X 304 = 2,736.    

Explanation: In standard algorithm we add same time of multiplying as 2 405 X 4 1,620 ,here 4 X 5 ones = 20 ones, we write 0 at ones place and take 2 to tens places then 4 x 0 ten = 0 ten + 2 tens = 2 tens, 4 X 4 hundreds = 16 hundreds, now we write 6 at hundreds place and 1 at thousand place as shown above, So 4 X 405 = 1,620.

Explanation: In standard algorithm we add same time of multiplying as 3 316 X 5 1,580 ,here 5 X 6 ones = 30 ones, we write 0 at ones place and take 3 to tens places then 5 x 1 ten = 5 tens + 3 tens = 8 tens, 5 X 3 hundreds = 15 hundreds, now we write 5 at hundreds place and 1 at thousand place as shown above, So 5 X 316 = 1,580.

Explanation: In standard algorithm we add same time of multiplying as 5,1 392 X 6 2,352 ,here 6 X 2 ones = 12 ones, we write 2 at ones place and take 1 to tens places then 6 x 9 tens = 54 tens + 1 tens = 55 tens, we write 5 at tens place and take another 5 to hundreds place, 6 X 3 hundreds = 18 hundreds,18 hundreds + 5 hundreds = 23 hundreds now we write 3 at hundreds place and 2 at thousands place as shown above, So 6 X 392 = 2,352.

Question 3. The product of 7 and 86 is ________. Answer: The product of 7 and 86 is 602,

Explanation: Give to find the product of 7 and 86 is 4 86 X 7 602 first we multiply 7 with 6 ones = 42 ones we write 2 at ones place and take 4 to tens place now 7 X 8 = 56 tens adding 4 tens we get 60 tens, So 7 X 86 = 602.

Question 4. 9 times as many as 457 is _________. Answer: 9 times as many as 457 is 4,113,

Explanation: Given to find 9 times as many as 457 is 6 457 X 9 4,113 ,here 9 X 7 ones = 63 ones, we write 3 at ones place and take 6 to tens places then 9 x 5 tens = 45 tens + 6 tens = 51 tens, we write 1 at tens place and take 5 to hundreds place, 9 X 4 hundreds = 36 hundreds, 36 hundreds + 5 hundreds = 41 hundreds now we write 1 at hundred place and 4 at thousands place as shown above, So 9 X 457 = 4,113.

Explanation: Given Jashawn wants to make 5 airplane propellers. He needs 18 centimeters of wood for each propeller. So number of centimeters of wood will he use is 5 X 18 cms = 4 18 X 5 90 here first we multiply 5 X 8 ones = 40 ones, we write 0 at ones place and take 4 to tens place then 5 X 1 ten = 5 tens+ 4 tens = 9 tens, So 5 X 18 = 90 centimeters, therefore, Jashawn needs 90 centimeters of wood to make 5 airplane propellers.

Question 6. One game system costs $238. How much will 4 game systems cost? Answer: 4 game systems will cost $952,

Explanation: Given one game system costs $238. So for 4 game systems it will cost 4 X $238 =  1, 3 $238 X  4 952 here first we multiply 4 X 8 ones = 32 ones, we write 2 at ones place and take 3 to tens places then 4 x 3 tens = 12 tens + 3 tens = 15 tens, we write 5 at tens place and take 1 to hundred place, 4 X 2 hundreds = 8 hundreds, now 8 hundreds + 1 hundred = 9 hundreds, So 5 X $238 = 952, therefore 4 game systems will cost $952.

Explanation: Given a small bag of chips weighs 48 grams. A large bag of chips weighs three times as much as the small bag means 3 X 48 grams= 2 48 X 3 144 here first we multiply 3 X 8 ones = 24 ones, we write 4 at ones place and take 2 to tens place then 3 X 4 ten = 12 tens+ 2 tens = 14 tens, we write 14 tens as 4 at tens place and 1 at hundreds place, So 3 X 48 grams = 144 grams, Now 7 large bags of chips weigh 7 X 144 grams =  3,2 144 X 7 1,008 here first we multiply 7 X 4 ones = 28 ones, we write 8 at ones place and take 2 to tens place then 7 X 4 ten = 28 tens+ 2 tens = 30 tens, we write 30 tens as 0 at tens place and other 3 at hundreds place,now 7 X 1 hundred = 7 hundred, 7 hundred + 3 hundreds = 10 hundreds we write 0 at hundreds place at 1 at thousands place, So 7 X 144 = 1,008 grams, therefore, 7 large bags of chips weigh 1,008 grams.

Eureka Math Grade 4 Module 3 Lesson 9 Exit Ticket Answer Key

Question 2. Morgan is 23 years old. Her grandfather is 4 times as old. How old is her grandfather? Answer: Morgan’s grandfather is 92 years old,

Explanation: Given Morgan is 23 years old and her grandfather is 4 times as old. So Morgan’s grandfather’s is 4 X 23 years = 1 23 X 4 92, here 4 X 3 ones = 12 ones, we write 2 at ones place and take 1 to tens places then 4 x 2 tens = 8 tens, 8 tens + 1 ten = 9 tens, So Morgan’s grandfather is 92 years old.

Eureka Math Grade 4 Module 3 Lesson 9 Homework Answer Key

Explanation: Given expression as 2 X 46 = Partial Products 46 X 2   12 +80 92 , Here we first write multiplication of 2 X 6 ones, then 2 X 4 tens then add as shown above 12 + 80 = 92, In standard algorithm we add same time of multiplying as 1 46 X 2 92 , here 2 X 6 ones= 12 ones we write 2 at ones place and take 1 at tens place then 2 x 4 at tens place = 8 tens add 1 ten to 8 tens we get 9 tens so 2 X 46 = 92.

Explanation: In standard algorithm we add same time of multiplying as 1 232 X 4 928 , here 4 X 2 ones = 8 ones then 4 x 3 tens = 12 tens we write 2 at tens place and take 1 to hundred place and 4 X 2 hundreds = 8 hundreds, now adding 1 hundred to 8 hundreds we get 9 hundreds , so 4 X 232 = 928.

Explanation: In standard algorithm we add same time of multiplying as 1 440 X 3 1,320 ,here 3 X 0 ones = 0 ones, we write 0 at ones place, 3 x 4 tens = 12 tens we write 2 at tens place and take 1 at hundreds place, 3 X 4 hundreds = 12 hundreds, 12 hundreds + 1 hundred = 13 hundreds ,now we write 3 at hundreds place and 1 at thousand place as shown above, So 3 X 440 = 1,320.

Question 3. What is the product of 8 and 54? Answer: The product of 8 and 54 is 432,

Explanation: Give to find the product of 8 and 54 is 3 54 X 8 432 first we multiply 8 with 4 ones = 32 ones we write 2 at ones place and take 3 to tens place now 8 X 5 = 40 tens adding 3 tens we get 43 tens, So 8 X 54 = 432.

Question 4. Isabel earned 350 points while she was playing Blasting Robot. Isabel’s mom earned 3 times as many points as Isabel. How many points did Isabel’s mom earn? Answer: Isabel’s mom earn 1050 points,

Explanation: Given Isabel earned 350 points while she was playing Blasting Robot. Isabel’s mom earned 3 times as many points as Isabel. So  points did Isabel’s mom earn is 3 X 350 = 1 350 X 3 1050, first we multiply 3 with 0 ones = 0 ones , 3 X 5 = 15 tens we write 15 tens as 5 at tens place and 1 at hundreds place and 3 X 3 = 9 hundreds + 1 hundred = 10 hundreds we write 10 hundreds as 0 at hundreds place and 1 at thousands place, So 3 X 350 = 1050. Therefore, Isabel’s mom earn 1050 points.

Question 5. To get enough money to go on a field trip, every student in a club has to raise $53 by selling chocolate bars. There are 9 students in the club. How much money does the club need to raise to go on the field trip? Answer: The club needs to raise $477 to go on the field trip,

Explanation: Given to get enough money to go on a field trip, every student in a club has to raise $53 by selling chocolate bars. There are 9 students in the club. So money does the club need to raise to go on the field trip is 9 X $53 = 2 $53 X 9 $477 first we multiply 9 with 3 ones = 27 ones , we write 7 at ones place and 2 at tens place, 9 X 5 tens= 45 tens, 45 tens + 2 tens = 47 tens we write 47 tens as 7 at tens place and 4 at hundreds place so 9 X $53 = $477, therefore,the club needs to raise $477 to go on the field trip.

Question 6. Mr. Meyers wants to order 4 tablets for his classroom. Each tablet costs $329. How much will all four tablets cost? Answer: Total cost for all four tablets are $1,316,

Explanation: Given Mr. Meyers wants to order 4 tablets for his classroom. Each tablet costs $329. So total cost for four tablets are 4 X $329 = 1,3 $329 X 4 $ 1,316 first we multiply 4 with 9 ones = 36 ones , we write 6 at ones place and 3 at tens place, 4 X 2 tens= 8 tens, 8 tens + 3 tens = 11 tens we write 11 tens as 1 at tens place and 1 at hundreds place, Now 4 X 3 hundreds = 12 hundreds, 12 hundreds + 1 hundred = 13 hundreds, we write 3 at hundreds place and 1 at thousands place, So 4 X $329 = $1,316.

Question 7. Amaya read 64 pages last week. Amaya’s older brother, Rogelio, read twice as many pages in the same amount of time. Their big sister, Elianna, is in high school and read 4 times as many pages as Rogelio did. How many pages did Elianna read last week? Answer: Elianna read 512 pages last week,

Explanation: Given Amaya read 64 pages last week. Amaya’s older brother, Rogelio, read twice as many pages in the same amount of time. So Rogelio reads 2 X 64 pages = 64 X 2 128 2 X 4 ones = 8  and 2 X 6 tens= 120, So 2 X 64 = 8 + 120 = 128, So Rogelio reads 128 pages, Their big sister, Elianna, is in high school and read 4 times as many pages as Rogelio did. So number of pages did Elianna read last week are 4 X 128 pages= 1, 3 128 x 4 512 here we first we multiply 4 with 8 ones = 32 ones , we write 2 at ones place and 3 at tens place, 4 X 2 tens= 8 tens, 8 tens + 3 tens = 11 tens we write 11 tens as 1 at tens place and 1 at hundreds place and 4 X 1 hundred = 4 hundreds, 4 hundreds +1 hundred = 5 hundreds, So 4 X 128 pages = 512 pages, therefore, Elianna read 512 pages last week.

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